Question

Number of seven letter words that can be formed by using the letters of the word $SUCCESS$ so that:
$\left(a\right)$ the two $C$ are together but no two $S$ are together be $k$.
$\left(b\right)$ no two $C$ and no two $S$ are together be $m$.
Find $m/k$ ?

Answer 1

$\left(a\right)$ Consider $CC$ as single object, $U,CC,E$ can be arranged in $3!$ ways
$\times U\times CC\times E\times$
Now the three $S$ are to be placed in the four available places.
Hence required no. of ways $=3!.{}^4{C}_3=24$
$\left(b\right)$ Let us first find the words in which no two $S$ are together
$\left(i\right)$ Arrange the remaining letters $=\frac{4!}{2!}=12$ ways
$\left(ii\right)\times U\times CC\times E\times$
There are five available places for three $SSS$.
Hence the total no. of ways no two $S$ together $=12\times {}^5{C}_3=120$
$\therefore$ Hence, no. of words having $CC$ separated and $SSS$ separated
$=120-24=96$
$\frac{96}{24}=4$